50.0 ml of 0.250M HCl is added to a flask containing 25.0 ml of 0.325M KOH. What

Question

50.0 ml of 0.250M HCl is added to a flask containing 25.0 ml of 0.325M KOH. What is the pH of the final solution? 0.561 1.23 7.00 12.76 When 50 mL of 0.1 M NaOH is added to 50 mL of 0.2 M solution of an acid HX, the pH of the resultant solution is 6. What is the K_a of HX? 1 times 10^-6 5 times 10^-7 2 times 10^-6 1 times 10^-8 For the titration of NH_3 with HCl, which of the following would not be a good indicator? phenolphthalein - changes color at pH 8 to 10 methyl orange - changes color at pH 3.5 to 5.0

Explanation / Answer

NaOH moles added = M x V = 0.1 x ( 50/1000) = 0.005

HX moles = M x V ( in L) = 0.2 x ( 50/1000) = 0.01

the rection we have is  

HX (aq) + NaOH (aq) <--> NaX (aq) + H2O ( l)

hence after adding NaOH   , resulting HX moles = 0.01-0.005 = 0.005

NAX moles formed = NaoH moles added = 0.005

volume of solution = 50+50 = 100 ml = 0.1L

[HX] = moles/ volume = 0.005/0.1 = 0.05 M

[NaX] = 0.005/0.1 = 0.05 M

we have a buffer now .i.e weak acid HX and its conjugate base NaX

pH of buffer is given by

pH = pka + log [ conjugate base] / [ acid]

6 = pka + log ( 0.05/0.05)

pka = 6

Ka = 10^ -pka = 10^ -6

hence Ka = 10^ -6

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